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ticbol
Posts:
116
Registered:
1/25/05


Re: Solving e^y  3e^(y) + 2 = 0
Posted:
Jan 30, 2006 5:30 AM


 wrote: > Please assist. I'm stuck with this: > > e^y  3e^(y) + 2 = 0 > e^y  3(e^y)^1 + 2 = 0 > Let x = e^y > x  3x^1 + 2 = 0 > I don't know what to do next (and am unsure if the above is correct)
 Here is one way.
e^y 3e^(y) +2 = 0 e^y 3/(e^y) +2 = 0 Clear the fraction, multiply both sides by e^y, (e^y)^2 3 +2e^y = 0 (e^y)^2 +2e^y 3 = 0 (e^y +3)(e^y 1) = 0
e^y +3 = 0 e^y = 3 Take the natural log of both sides, y*ln(e) = ln(3) y = ln(3) cannot be, so reject this.
e^y 1 = 0 e^y = 1 y*ln(e) = ln(1) y = 0 answer.
Check, e^y 3e^(y) +2 = 0 e^0 3e^(0) +2 =? 0 1 3(1) +2 =? 0 1 3 +2 =? 0 0 = 0 Yes, so, OK.
 Is negative zero, zero? If you do not like that, then 3e^(y) = 3/(e^y) > 3/(e^0) = 3/1 = 3.



